Calculus with Applications (10th Edition)

Published by Pearson
ISBN 10: 0321749006
ISBN 13: 978-0-32174-900-0

Chapter 4 - Calculating the Derivative - 4.1 Techniques for Finding Derivatives - 4.1 Exercises - Page 208: 37



Work Step by Step

Horizontal lines have slope equal to 0, so to determine the values of $x$ where the tangent line is horizontal, we set the derivative of $f$ equal to $0$ and solve for $x$. Using the Power Rule, the derivative of $$f(x)= 2x^3 +9x^2-60x+4$$ is $$f'(x) = 6x^2+18x-60.$$ We set the derivative $f'(x)$ equal to 0 and solve for $x$: $$f'(x) = 6x^2+18x-60=0.$$ Factor out a 6 from the left side: $$6( x^2+3x-10)=0.$$ Dividing both sides of the equation by 6 gives: $$ x^2+3x-10=0.$$ We factor the quadratic equation to obtain: $$(x+5)(x-2)=0.$$ Setting each factor equal to $0$ and solving for $x$, we get that $x=-5$ or $x=2$.
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